A box contains 12 transistors, 4 of which are defective. If 4 are selected at random, find the probability of the statements below. a. All are defective b. None are defective a. The probability is (Type a fraction. Simplify your answer.) b. The probability is (Type a fraction. Simplify your answer.)

Step 1 ：The problem is asking for the probability of selecting 4 defective transistors out of 12, and the probability of selecting 4 non-defective transistors out of 12.

Step 2 ：To solve this, we can use the combination formula which is \(nCr = \frac{n!}{r!(n-r)!}\), where n is the total number of items, r is the number of items to choose, and '!' denotes factorial.

Step 3 ：For the first part, n is 4 (the number of defective transistors) and r is 4 (the number of transistors we are choosing).

Step 4 ：For the second part, n is 8 (the number of non-defective transistors) and r is 4.

Step 5 ：The total number of ways to choose 4 transistors out of 12 is \(12C4\), which is 495.

Step 6 ：The number of ways to choose 4 defective transistors out of 4 is \(4C4\), which is 1.

Step 7 ：The number of ways to choose 4 non-defective transistors out of 8 is \(8C4\), which is 70.

Step 8 ：The probability of all 4 transistors being defective is \(\frac{1}{495}\), which simplifies to 0.002.

Step 9 ：The probability of none of the 4 transistors being defective is \(\frac{70}{495}\), which simplifies to 0.141.

Step 10 ：Final Answer: The probability of all 4 transistors being defective is \(\boxed{0.002}\). The probability of none of the 4 transistors being defective is \(\boxed{0.141}\).